Q: The instantaneous gas discharge of the system and the C-E

[tracker890 Source h]

Homework Statement

Confusing in the conservation of mass.

Relevant Equations

Continuity Equation

Q： Why is air expelled from the system, yet the system's mass remains unchanged? Isn't mass related to volume?

 

[Frabjous]

The system includes the expelled air.

 

[tracker890 Source h]

If the system includes air output, how should the cross-section be determined?
In a closed system, isn't it the case that fluids cannot cross the system boundaries? However, the following equation is indeed correct.

The system includes the expelled air.

 

[erobz]

View attachment 331684
If the system includes air output, how should the cross-section be determined?
View attachment 331685

The system is the total mass. It's not changing in time. The tank is the control volume. The portion of the systems mass inside the control volume is changing in time, so you get:

$$ \frac{d}{dt} \int_{cv} \rho ~d V \llap{-} = - \int_{cs} \rho \vec{V} \cdot d \vec{A} $$

which reduces to ( uniformily distributed properties):

$$ \frac{d}{dt} m_{cv} = - \rho_e A_e V_e $$

 

[berkeman]

Q： Why is air expelled from the system, yet the system's mass remains unchanged? Isn't mass related to volume?

Could you please explain why you are taking the time derivative of a 4-dimensional volume integral?

 

[erobz]

Could you please explain why you are taking the time derivative of a 4-dimensional volume integral?

View attachment 331686

I was going to mention there are one too many integral signs.

 

[tracker890 Source h]

The system is the total mass. It's not changing in time. The tank is the control volume. The portion of the systems mass inside the control volume is changing in time, so you get:

$$ \frac{d}{dt} \int_{cv} \rho ~d V \llap{-} = - \int_{cs} \rho \vec{V} \cdot d \vec{A} $$

which reduces to ( uniformily distributed properties):

$$ \frac{d}{dt} m_{cv} = - \rho_e A_e V_e $$

So, my thoughts are as follows, is this correct?

 

[tracker890 Source h]

Could you please explain why you are taking the time derivative of a 4-dimensional volume integral?

View attachment 331686

Accidentally typed the wrong characters, it has been corrected.

 

[erobz]

So, my thoughts are as follows, is this correct?
View attachment 331687

Yeah, that's how you are to look at it. That first derivative being non-zero is not something you are going to encounter when it comes to mass as the property in R.T.T.

In essence the system is not the control volume, its "the stuff" moving through the control volume, be it mass, momentum , or energy. The "system" enters the control volume, exits the control volume, and/or accumulates within the control volume.

 

[tracker890 Source h]

Yeah, that's how you are to look at it. That first derivative being non-zero is not something you are going to encounter when it comes to mass as the property in R.T.T.

In essence the system is not the control volume, its "the stuff" moving through the control volume, be it mass, momentum , or energy. The "system" enters the control volume, exits the control volume, and/or accumulates within the control volume.

Thank you for the detailed and patient explanation. ^^